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Matrices (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)4 Marks
Question
If $\left[\begin{array}{cc}2 a+b & 3 a-b \\ c+2 d & 2 c-d\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ 4 & -1\end{array}\right]$, find $a , b , c$ and $d$.
✓

Answer

$
\left[\begin{array}{cc}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]
$
By equality of matrices,
$
\begin{aligned}
& 2 a+b=2 \ldots . .(1) \\
& 3 a-b=3 \ldots \ldots .(2) \\
& c+2 d=4 \ldots \ldots .(3) \\
& 2 c-d=-1 \ldots \ldots .(4)
\end{aligned}
$
Adding (1) and (2), we get
$
\begin{aligned}
& 5 a=5 \\
& \therefore a=1
\end{aligned}
$
Substituting $a =1$ in (1), we get
$
\begin{aligned}
& 2(1)+b=2 \\
& \therefore b=0
\end{aligned}
$
Multiplying equation (4) by 2 , we get
$
4 c -2 d =-2 \ldots \ldots \text { (5) }
$
Adding (3) and (5), we get
$
\begin{aligned}
& 5 c =2 \\
& \therefore c =\frac{2}{5}
\end{aligned}
$
Substituting $c=\frac{2}{5}$ in (4), we get
$
\begin{aligned}
& 2\left(\frac{2}{5}\right)-d=-1 \\
& \therefore d=\frac{4}{5}+1=\frac{9}{5}
\end{aligned}
$
Hence, $a=1, b=0, c=\frac{2}{5}$ and $d=\frac{9}{5}$.

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